3 text{ mole} of a mixture of FeSO_4 and Fe_2 | vedclass

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(D) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and reduces to $Mn^{2+}$,involving a change in oxidation state of $+5$ ($n$-factor $= 5$).

Vedclass · Accepted Answer

$\frac{1}{3}$

Vedclass · Answer

(D) In an acidic medium,$KMnO_4$ acts as an oxidizing agent and reduces to $Mn^{2+}$,involving a change in oxidation state of $+5$ ($n$-factor $= 5$).$FeSO_4$ contains $Fe^{2+}$,which gets oxidized to $Fe^{3+}$ ($n$-factor $= 1$).$Fe_2(SO_4)_3$ contains $Fe^{3+}$,which cannot be further oxidized by $KMnO_4$.Equating the milliequivalents (meq) of the reducing agent $(FeSO_4)$ and the oxidizing agent $(KMnO_4)$:$	ext{meq of } FeSO_4 = 	ext{meq of } KMnO_4$$	ext{moles of } FeSO_4 	imes 1 = 	ext{Molarity} 	imes 	ext{Volume (in mL)} 	imes n	ext{-factor}$$	ext{moles of } FeSO_4 = 2 	imes 100 	imes 5 / 1000 = 1 	ext{ mole}$.Total moles of mixture $= 3 	ext{ moles}$.$	ext{Mole fraction of } FeSO_4 = \frac{	ext{moles of } FeSO_4}{	ext{Total moles}} = \frac{1}{3}$.

$3 \text{ mole}$ of a mixture of $FeSO_4$ and $Fe_2(SO_4)_3$ required $100 \text{ mL}$ of $2 \text{ M}$ $KMnO_4$ solution in acidic medium. Hence,the mole fraction of $FeSO_4$ in the mixture is:

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